Question:
If \( \int_{4}^{9} \dfrac{x-\sqrt{x}}{x(x-1)}\, \mathrm{d}x=\alpha \), then \( \mathrm{e}^{\alpha} \) is equal to :
If \( \int_{4}^{9} \dfrac{x-\sqrt{x}}{x(x-1)}\, \mathrm{d}x=\alpha \), then \( \mathrm{e}^{\alpha} \) is equal to :
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Factor \( x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1) \) and \( x-1=(\sqrt{x}-1)(\sqrt{x}+1) \), then substitute \( u=\sqrt{x} \).
Updated On: Nov 12, 2026
- \( 9/4 \)
- \( 16/9 \)
- \( 4/9 \)
- \( 4/3 \)
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The Correct Option is B
Solution and Explanation
\( x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1) \) and \( x-1=(\sqrt{x}-1)(\sqrt{x}+1) \), so the integrand is \( \dfrac{\sqrt{x}}{x(\sqrt{x}+1)}=\dfrac{1}{\sqrt{x}(\sqrt{x}+1)} \).
Put \( u=\sqrt{x} \), \( dx=2u\,du \); limits \( x=4\to u=2 \), \( x=9\to u=3 \).
\( \alpha=\int_2^3\dfrac{2u\,du}{u(u+1)}=2\int_2^3\dfrac{du}{u+1}=2\left[\ln(u+1)\right]_2^3=2\ln\dfrac{4}{3}=\ln\dfrac{16}{9} \).
So \( e^{\alpha}=\dfrac{16}{9} \).
Answer: (2) 16/9
Put \( u=\sqrt{x} \), \( dx=2u\,du \); limits \( x=4\to u=2 \), \( x=9\to u=3 \).
\( \alpha=\int_2^3\dfrac{2u\,du}{u(u+1)}=2\int_2^3\dfrac{du}{u+1}=2\left[\ln(u+1)\right]_2^3=2\ln\dfrac{4}{3}=\ln\dfrac{16}{9} \).
So \( e^{\alpha}=\dfrac{16}{9} \).
Answer: (2) 16/9
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